\subsection{Normal subgroups}
How and when does it make sense to divide one group by another?
\begin{definition}
	An subgroup \(N\) of a group \(G\) is \textit{normal} if \(\forall g \in G, gN = Ng\).
	We write \(N \trianglelefteq G\).
\end{definition}
The following equivalent definitions hold:
\begin{itemize}
	\item \(\forall g \in G, gN = Ng\)
	\item \(\forall g \in G, \forall n \in N, g^{-1}ng \in N\)
	\item \(\forall g \in G, g^{-1}Ng = N\)
\end{itemize}
\begin{proof}
	The first case is the definition.
	For the second case, clearly (from the first definition) \(ng \in gN\).
	So multiplying on the left by \(g^{-1}\), we have \(g^{-1}ng \in N\) as required.
	For the third case, we can simply multiply the first definition on the left by \(g^{-1}\).
	Note that these multiplications are distributed over each element in the coset: \(a(bC) = \{ abc : c \in C \}\).
\end{proof}

\begin{enumerate}
	\item \(\{ e \}\) and \(G\) are normal subgroups of \(G\).
	\item \(n\mathbb Z \trianglelefteq \mathbb Z\).
	      \(\forall a \in \mathbb Z\), we have \(a + n\mathbb Z = \{ a + nk : k \in \mathbb Z \} = \{ nk + a : k \in \mathbb Z \} = n\mathbb Z + a\).
	\item \(A_3 \trianglelefteq S_3\).
	      \begin{itemize}
		      \item \(eA_3 = A_3 = A_3 e\)
		      \item \((1\ 2\ 3)A_3 = A_3 = A_3(1\ 2\ 3)\)
		      \item \((1\ 3\ 2)A_3 = A_3 = A_3(1\ 3\ 2)\)
		      \item \((1\ 2)A_3 = \{ (1\ 2), (2\ 3), (1\ 3) \} = A_3(1\ 2)\)
	      \end{itemize}
	      and so on.
\end{enumerate}

\begin{proposition}
	\begin{enumerate}
		\item Any subgroup of an abelian group is normal.
		\item Any subgroup of index 2 is normal.
	\end{enumerate}
\end{proposition}
\begin{proof}
	\begin{enumerate}
		\item If \(G\) is abelian, then \(\forall g \in G, \forall n \in N, g^{-1}ng = n \in N\) which is stronger than required.
		\item If \(H \leq G\) with \(\abs{G : H} = 2\), then there are only 2 cosets.
		      \(H = eH = He\) is one of the two cosets.
		      Since cosets are disjoint, the other coset must be \(G \setminus H\).
		      This is true for both left and right cosets.
		      So the other left and right cosets must be equal, so \(H\) is normal.
	\end{enumerate}
\end{proof}

\begin{proposition}
	If \(\varphi: G \to H\) is a homomorphism, then \(\ker \varphi \trianglelefteq G\).
\end{proposition}
\begin{proof}
	We already know \(\ker \varphi\) is a subgroup of \(G\).
	Now we must check it is normal.
	Given some \(k \in \ker \varphi, g \in G\), we want to show that \(g^{-1} k g \in \ker \varphi\).
	We have \(\varphi(g^{-1} k g) = \varphi(g^{-1}) \varphi(k) \varphi(g) = \varphi(g^{-1}) e \varphi(g) = \varphi(g^{-1}g) = \varphi(e) = e\) so \(g^{-1} k g \in \ker \varphi\) as required.
\end{proof}
In fact, we will show later that normal subgroups are exactly kernels of homomorphisms and nothing else.

Here is now a less formal explanation of this theorem and its consequences.
Consider some subgroup \(K \leq G\).
There may be some property \(P\) that is true for every element of \(K\) and false for every other element of \(G\).
Then certainly, for example, given \(k_1, k_2 \in K\), we know that \(k_1k_2\) has the same property as it is within \(K\).
As another example, let \(k \in K\) and let \(g \in G \setminus K\).
Then \(kg\) does not have this property, as \(kg \notin K\).

We can encapsulate this behaviour by making a homomorphism from the whole group \(G\) to some other group---it \textit{doesn't matter where we end up}, just as long as anything with this particular property maps to the new group's identity element.
Let \(\varphi: G \to H\), where \(H\) is some group that we don't really care about (apart from the identity).
This means that any element of \(K\), i.e.\ any element with property \(P\), is mapped to \(e_H\).
By the laws of homomorphisms, any product of \(k \in K\) with \(g \in G \setminus K\) does not give the identity element, so it does not have this property!
This is exactly the behaviour we wanted.

If we can find such a homomorphism, then \(K\) is the kernel of this homomorphism.
Again, the image of this homomorphism is essentially irrelevant; all we care about is which elements map to the identity.
Now, note that by the laws of homomorphisms, given some element \(g \in G\) and \(k \in K\), \(\varphi(g^{-1}kg) = \varphi(g^{-1})\varphi(k)\varphi(g)\).
But since \(k\) has this desired property, the \(\varphi(k)\) term vanishes.
So we're left with the identity element.
This gives us the result that \(g^{-1}kg\) must be an element of \(K\), so it must have property \(P\).
This is a definition for a normal subgroup, so \(K\) must be normal in order for us to be able to find such a homomorphism \(\varphi\).

As another small aside, a normal subgroup in this context essentially means this: given some element \(k\) with property \(P\), the property is preserved when surrounding \(k\) with inverses.
This is just a `translation' of a definition of a normal subgroup: \(g^{-1}kg \in K\).

\begin{enumerate}
	\item \(SL_n(\mathbb R) \trianglelefteq GL_n(\mathbb R)\), where \(GL_n(\mathbb R)\) is the group of invertible matrices of dimension \(n\), and where \(SL_n(\mathbb R)\) is the group of matrices of determinant 1.
	      This is because \(\det: GL_n(\mathbb R) \to \mathbb R^*\), and \(SL_n(\mathbb R) = \ker (\det)\).
	\item \(A_n \trianglelefteq S_n\) as \(A_n\) is the kernel of the sign homomorphism.
	      Alternatively, it is an index 2 subgroup so it must be normal.
	\item \(n\mathbb Z \trianglelefteq \mathbb Z\) as the kernel of \(\varphi: \mathbb Z \to \mathbb Z_n\), where \(\varphi(k) = k \text{ mod } n\), or since \(\mathbb Z\) is abelian.
\end{enumerate}
With this notion of normal subgroups, we can make some progress into categorising small groups.
\begin{proposition}
	If \(\abs{G} = 6\), then \(G \cong C_6\) or \(G \cong D_6\).
\end{proposition}
\begin{proof}
	By Lagrange's Theorem, the possible element orders are 1 (only the identity), 2, 3, 6.
	\begin{itemize}
		\item If there is an element \(g\) of order 6, then \(G = \genset g \cong C_6\).
		\item Otherwise, (again by question 7 on example sheet 1) there must be an element of the group not of order 2, because if we just had elements of order 2 then \(\abs{G}\)
		      would have to be a power of 2.
		      So there is an element \(r\) of order 3, so \(\abs{\genset{r}} = 3\), and by Lagrange's Theorem \(\abs{G} = 6 = \abs{G:\genset{r}} \cdot \abs{\genset{r}}\), so \(\abs{G:\genset{r}} = 2\).
		      So \(\genset{r} \trianglelefteq G\).
		      There must also be an element \(s\) of order 2, since \(\abs{G}\) is even (by question 8 from example sheet 1).

		      So, what can \(s^{-1} r s\) be?
		      Because \(\genset{r}\) is normal, then \(s^{-1} r s \in \genset{r}\).
		      So it is either \(e\), \(r\) or \(r^2\).
		      \begin{itemize}
			      \item If \(s^{-1}rs = e\) then \(r = e\) \contradiction{}
			      \item If \(s^{-1}rs = r\) then \(sr=rs\), and so \(sr\) has order \(\LCM(\ord s, \ord r) = \LCM(2, 3) = 6\) \contradiction{}
			      \item So \(s^{-1}rs = r^2\), then \(G = \genset{r, s}\) with \(r^3 = s^2 = e\) and \(sr = r^2 s = r^{-1}s\), which are the defining features of \(D_6\).
		      \end{itemize}
	\end{itemize}
\end{proof}

\subsection{Motivation for quotients}
Let us consider \(n\mathbb Z \trianglelefteq \mathbb Z\).
The cosets are \(0+n\mathbb Z\), \(1 + n\mathbb Z, \cdots, (n-1) + n\mathbb Z\).
These cosets, although they are subsets of \(\mathbb Z\), behave a lot like the elements of the group \(\mathbb Z_n\).
For example, if we try to define addition between the cosets:
\[
	(k + n \mathbb Z) + (m + n \mathbb Z) \coloneq (k+m) + n \mathbb Z
\]
which acts like addition modulo \(n\mathbb Z\).
For a general subgroup \(H \leq G\), we could try to do the same.
\[
	g_1 H \cdot g_2 H \coloneq g_1 g_2 H
\]
But we can write the cosets on the left hand side in many ways, as the representation is dependent on the choice of representative for each coset, so this multiplication may not be well defined.
We can guarantee that it is well defined (so that we can turn the set of cosets into a group) by ensuring that
\[
	g_1' H = g_1 H;\; g_2' H = g_2 H \implies g_1'g_2'H = g_1g_2H
\]
If \(g_1' H = g_1 H;\; g_2' H\), then \(g_1' = g_1h_1\) and \(g_2' = g_2h_2\) for some \(h_1 h_2 \in H\).
So
\[
	g_1'g_2'H = g_1h_1g_2\underbrace{h_2H}_{\mathclap{h_2H = \{ h_2h : h \in H \} = H}}
\]
So in order to get \(g_1' g_2' H = g_1 g_2 H\), we need \(g_1 h_1 g_2 H = g_1 g_2 H\) for any elements \(g_1, g_2, h_1\) that we choose.
Therefore:
\begin{align*}
	g_1h_1g_2H         & = g_1g_2 H \\
	g_2^{-1} h_1 h_2 H & = H        \\
	\text{or } g_2^{-1} h_1 g_2 \in H\;(\forall g_2 \in G, h_1 \in H)
\end{align*}
This is an equivalent condition for the subgroup to be normal.

\subsection{Quotients}
\begin{proposition}
	Let \(N \trianglelefteq G\).
	The set of (left) cosets of \(N\) in \(G\) forms a group under the operation \(g_1N\cdot g_2N = g_1g_2N\).
\end{proposition}
\begin{proof}
	The group operation is well defined as shown above.
	We now show the group axioms hold.
	\begin{itemize}
		\item (closure) If \(g_1N, g_2N\) are cosets, then \(g_1g_2N\) is also a coset.
		\item (identity) \(eN = N\)
		\item (inverses) \((gN)^{-1} = g^{-1}N\)
		\item (associativity) Follows from the associativity of \(G\): \((g_1 N \cdot g_2 N)\cdot g_3 N = g_1g_2 N \cdot g_3 N = g_1g_2g_3 N = g_1 N \cdot g_2g_3 N = g_1 N \cdot (g_2 N \cdot g_3 N)\)
	\end{itemize}
\end{proof}

\begin{definition}
	If \(N \trianglelefteq G\), the group of (left) cosets of \(N\) in \(G\) is called the quotient group of \(G\) by \(N\), written \(\faktor{G}{N}\).
\end{definition}

This is a nice way of thinking about quotient groups.
Imagine you have a group \(N\) of some distinct objects \(n_1, n_2, n_3\) and so on.
Imagine lining them all up in a row of length \(\abs{N}\).
Then the cosets of \(N\) in \(G\) can be thought of as `translated copies' of \(N\).
For example, let the cosets of \(N\) in \(G\) be \(N, g_1N, g_2N\) and so forth.
Now, picture these cosets as copies of \(N\), translated downwards on the page, so that they are like multiple rows, and that therefore there we have a grid containing all elements of \(G\).
Now, we have formed a rectangle of area \(\abs{G}\) out of \(\abs{N}\) columns and \(c\) rows, where \(c\) is the amount of `copies' of \(N\).
Therefore, \(c = \frac{\abs{G}}{\abs{N}}\), as the area of a rectangle is width multiplied by height.

Now, given some element in one of the cosets (i.e.\ in \(G\)) we can do some transformation \(g\) to take us to another element.
But because we made cosets out of a normal subgroup, multiplying by \(g\) is the same as swapping some of the rows, then maybe moving around the order of the elements in each row.
It keeps the identity of each row consistent---all elements in a given row are transformed to the same output row.
Remember that the word `row' basically means `coset'.

This means that we can basically forget about the individual elements in these cosets, all that we really care about is how the rows are swapped with each other under a given transformation.
Note, the quotient of 5 in 100 is 20, because there are 20 copies of 5 in 100.
So the quotient group of \(N\) in \(G\) is just all the copies of \(N\) in \(G\).
The group operation is simply the transformation of rows.
If we're talking about \(\faktor{G}{N}\), ask the question: `how do the copies of \(N\) in \(G\) behave'?

\subsection{Examples and properties}
\begin{enumerate}
	\item The cosets of \(n\mathbb Z\) in \(\mathbb Z\) give a group that behaves exactly like \(\mathbb Z_n\).
	      We write \(\faktor{\mathbb Z}{n\mathbb Z} \cong \mathbb Z_n\).
	      In fact, these are the only quotients of \(\mathbb Z\), as these are the only subgroups of \(\mathbb Z\).
	\item \(A_3 \trianglelefteq S_3\) gives \(\faktor{S_3}{A_3}\) which has only two elements since \(\abs{S_3 : A_3} = 2\), so it is isomorphic to \(C_2\).
	      Note that in general, \(\abs{G:N} = \abs{\faktor{G}{N}}\).
	\item If \(G = H \times K\), then both \(H\) and \(K\) are normal subgroups of \(G\).
	      We have \(\faktor{G}{H} \cong K\) and \(\faktor{G}{K} \cong H\) % (TODO: proof as exercise).
	\item Consider \(N \coloneq \genset{r^2} \trianglelefteq D_8\).
	      We can check that it is normal by trying \(r^{-1}r^2r^{-1} \in N\), and also \(s^{-1}r^2s = r^{-2} = r^2 \in N\).
	      Since \(\genset{r, s} = D_8\), and the generators obey this normal subgroup relation, it follows that \(g^{-1}ng\) for all \(g \in D_8\).
	      We know \(\abs{N} = 2\), so \(\abs{\faktor{D_8}{N}} = \abs{D_8 : N} = \frac{\abs{D_8}}{\abs{N}}\) by Lagrange's Theorem.
	      So \(\abs{\faktor{D_8}{N}} = 4\).
	      We know that any group of order 4 is isomorphic either to \(C_4\) or \(C_2 \times C_2\).
	      We can check that the cosets are \(\faktor{D_8}{N} = \{ N, sN, rN, srN \}\) which does not contain an element of order 4, so it is isomorphic to \(C_2 \times C_2\).
\end{enumerate}
We now show a non-example using the subgroup \(H \coloneq \genset{(1\ 2)} \leq S_3\) which is not normal, e.g.
\((1\ 2\ 3) H \neq H (1\ 2\ 3)\).
The cosets are
\[
	H;\quad (1\ 2\ 3)H = \{(1\ 2\ 3), (1\ 3)\};\quad (1\ 3\ 2)H = \{(1\ 3\ 2), (2\ 3)\}
\]
Attempting a multiplication gives
\[
	(1\ 2\ 3)H \cdot (1\ 3\ 2) H = (1\ 2\ 3) (1\ 3\ 2) H = H
\]
but using a different coset representative,
\[
	(1\ 3)H \cdot (1\ 3\ 2) H = (1\ 3)(1\ 3\ 2) H = (2\ 3)H \neq H
\]
so the multiplication is not well defined so we cannot form the quotient.

\begin{itemize}
	\item We can check that certain properties are inherited into quotient groups from the original group, such as being abelian and being finite.
	\item Quotients are not subgroups of the original group.
	      They are associated with tha original group in a very different way to subgroups---in general, a coset may not even be isomorphic to a subgroup in the group.
	      The example with direct products above was an example that is not true in general.
	\item With normality, we need to specify in which group the subgroup is normal.
	      For example, if \(K \leq N \leq G\), with \(K \trianglelefteq N\).
	      This does not imply that \(K \trianglelefteq G\), this would require that \(g^{-1}Kg = K\) for all elements \(g\) in \(G\), but we only have that \(n^{-1}Kn = K\) for all elements \(n\) in \(N\), which is a weaker condition.
	      Normality is not transitive---for example, \(K \trianglelefteq N \trianglelefteq G\) does not imply \(K \trianglelefteq G\).
	\item However, if \(N \leq H \leq G\) and \(N \trianglelefteq G\), then the weaker condition \(N \trianglelefteq H\) is true.
\end{itemize}

\begin{theorem}
	Given \(N \trianglelefteq G\), the function \(\pi: G \to \faktor{G}{N}\), \(\pi(g) = gN\) is a surjective homomorphism called the quotient map.
	We have \(\ker \pi = N\).
\end{theorem}
\begin{proof}
	We prove that \(\pi\) is a homomorphism.
	\(\pi(g)\pi(h) = gN \cdot hN = (gh)N = \pi(gh)\) as required.
	Clearly it is surjective we we can create all possible cosets by applying the \(\pi\) function to a coset representative.
	Also, \(\pi(g) = gN = N\) if and only if \(g \in N\), so \(\ker \pi = N\).
\end{proof}
Therefore, normal subgroups are exactly kernels of homomorphisms.
Using the idea of `properties' for normal subgroups above, the property in question here is `belonging to \(N\)'.
Any element of \(N\) is in the coset \(N\), which is the identity coset of \(\faktor{G}{N}\).
Essentially, the first row of this quotient `grid' (as described above) is \(N\), which acts as the identity element in the \(\faktor{G}{N}\) quotient group.
